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NCERT solutions for Class 10th Maths Chapter 1:Real Numbers





In this page we have NCERT solutions for Class 10th Maths Chapter 1:Real Numbers for EXERCISE 1 . Hope you like them and do not forget to like , social_share and comment at the end of the page.
1. Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
4. Use Euclid’s division lemma to show that the square of any positive integer is either ofthe form 3m or 3m + 1 for some integer m.
[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]
5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form  9m, 9m + 1 or 9m + 8
Solution 1
We know two things from the chapter
Euclid’s Division Lemma
For a and b any two positive integer, we can always find unique integer q and r such that
a=bq + r    ,   0 ≤ r ≤ b
If r =0, then b is divisor of a.
HCF (Highest common factor)
HCF of two positive integers can be find using the Euclid’s Division Lemma algorithm
We know that for any two integers a,b. we can write following expression
a=bq + r    ,  0 ≤ r ≤ b
If r=0 ,then
HCF( a,b) =b
If r≠0 , then
HCF ( a,b)  = HCF ( b,r)
Again expressing the integer b,r in Euclid’s Division Lemma, we get
  b=pr + r1
HCF ( b,r)=HCF ( r,r1)
Similarly successive Euclid ‘s division can be written until we get the remainder zero, the divisor at that point is called the HCF of the a  and b
We will using the same in this question
i)  We should always start with Larger number
So in this case  the larger number is 225
Now we can write 135 and 225 in Euclid division algorithm
225=135 X 1 +90
Now HCF of numbers  (225,135) = HCF ( 135,90)
Again writing 135,90 is Euclid division formula
135=90X1 +45
Now HCF (135,90)= HCF (90,45)
Again writing 90,45 is Euclid division formula
90=45X2 +0
Now r =0,  45 is HCF (90,45)
 45 is HCF ( 225,135)
ii) We should always start with Larger number
So in this case  the larger number is 38220
Now we can write 38220 and 196   in Euclid division algorithm
38220=196 X 195 +0
 Now r =0,  196 is HCF (196,38220)
iii) We should always start with Larger number
So in this case  the larger number is 867
Now we can write 867 and 255 in Euclid division algorithm
867=255 X 3 +102
Now HCF of numbers  (867,255) = HCF ( 255,102)
Again writing 255,102 is Euclid division formula
255=102X2 +51
Now HCF (255,102)= HCF (102,51)
Again writing 102,51 is Euclid division formula
102=51X2 +0
Now r =0,  51 is HCF (102,51)
 51 is HCF ( 867,255)
Solution 2:
Euclid’s Division Lemma
For a and b any two positive integer, we can always find unique integer q and r such that
     a=bq + r    ,      0 ≤ r < b
Now on putting b=6  ,we get
a=6q+ r     , 0 ≤ r < 6
a=6q   , This is a even number
a=6q+1  , This is a odd number
a=6q+2  ,This is an even number as 6 and 2 are divisible by 2
a=6q+3, it is not divisible by 2
a=6q+4, it is divisible by 2
a=6q+5 , it is not divisible by 2
So  6q,6q+2,6q +4 are even number
6q+1,6q+3,6q+5  are odd numbers
Solution 3:
According to the questions, we need to  find the maximum  number of column
Maximum number of column is  the HCF of number 32 and 616
So question has reduced to finding the HCF of 32 and 616 using Euclid division algorithm
We should always start with Larger number
So in this case  the larger number is 616
Now we can write 616 and 32 in Euclid division algorithm
616=32 X19 +8
Now HCF of numbers  (616,32) = HCF ( 32,8)
Again writing 32,8 is Euclid division formula
32=8X4+0
As r=0,   8 is the HCF of 616 and 32
Solution 4:
Euclid’s Division Lemma
For a and b any two positive integer, we can always find unique integer q and r such that
  a=bq + r    ,      0 ≤ r < b
Now on putting b=3  ,we get
a=3q+ r     , 0 ≤ r < 3
a=3q   , a2=9q2  a2 =3m  where m=3q2
a=3q+1  , a2=9q2 +6q+1=3(3q2+2q) +1 =3m+1 where m=3q2+2q
a=3q+2  , a2=9q2 +12q+4=3(3q2+4q+1) +1 =3m+1 where m=3q2+4q+1
Solution 5:
Euclid’s Division Lemma
For a and b any two positive integer, we can always find unique integer q and r such that
a=bq + r    ,      0 ≤ r < b
Now on putting b=3  ,we get
a=3q+ r     , 0 ≤ r < 3
a=3q   , a3=27q3  a3 =9m  where m=3q3
a=3q+1  , a3=27q3 +27q2 +9q +1=9(3q3+3q+q) +1 =9m+1 where m=3q3+3q+q
a=3q+2  , a2=27q3 +54q2 +36q +8=9(3q3+6q+4q) +1 =9m+1 where m=3q3+6q+4q
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